\(\int \frac {\log (1+x)}{x \sqrt {1+\sqrt {1+x}}} \, dx\) [7]

Optimal result

Integrand size = 21, antiderivative size = 291 \[ \int \frac {\log (1+x)}{x \sqrt {1+\sqrt {1+x}}} \, dx=-8 \text {arctanh}\left (\sqrt {1+\sqrt {1+x}}\right )-\frac {2 \log (1+x)}{\sqrt {1+\sqrt {1+x}}}-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\sqrt {1+x}}}{\sqrt {2}}\right ) \log (1+x)+2 \sqrt {2} \text {arctanh}\left (\frac {1}{\sqrt {2}}\right ) \log \left (1-\sqrt {1+\sqrt {1+x}}\right )-2 \sqrt {2} \text {arctanh}\left (\frac {1}{\sqrt {2}}\right ) \log \left (1+\sqrt {1+\sqrt {1+x}}\right )+\sqrt {2} \operatorname {PolyLog}\left (2,-\frac {\sqrt {2} \left (1-\sqrt {1+\sqrt {1+x}}\right )}{2-\sqrt {2}}\right )-\sqrt {2} \operatorname {PolyLog}\left (2,\frac {\sqrt {2} \left (1-\sqrt {1+\sqrt {1+x}}\right )}{2+\sqrt {2}}\right )-\sqrt {2} \operatorname {PolyLog}\left (2,-\frac {\sqrt {2} \left (1+\sqrt {1+\sqrt {1+x}}\right )}{2-\sqrt {2}}\right )+\sqrt {2} \operatorname {PolyLog}\left (2,\frac {\sqrt {2} \left (1+\sqrt {1+\sqrt {1+x}}\right )}{2+\sqrt {2}}\right ) \]

[Out]

-8*arctanh((1+(1+x)^(1/2))^(1/2))-arctanh(1/2*(1+(1+x)^(1/2))^(1/2)*2^(1/2))*ln(1+x)*2^(1/2)+2*arctanh(1/2*2^(
1/2))*ln(1-(1+(1+x)^(1/2))^(1/2))*2^(1/2)-2*arctanh(1/2*2^(1/2))*ln(1+(1+(1+x)^(1/2))^(1/2))*2^(1/2)+polylog(2
,-2^(1/2)*(1-(1+(1+x)^(1/2))^(1/2))/(2-2^(1/2)))*2^(1/2)-polylog(2,2^(1/2)*(1-(1+(1+x)^(1/2))^(1/2))/(2+2^(1/2
)))*2^(1/2)-polylog(2,-2^(1/2)*(1+(1+(1+x)^(1/2))^(1/2))/(2-2^(1/2)))*2^(1/2)+polylog(2,2^(1/2)*(1+(1+(1+x)^(1
/2))^(1/2))/(2+2^(1/2)))*2^(1/2)-2*ln(1+x)/(1+(1+x)^(1/2))^(1/2)

Rubi [F]

\[ \int \frac {\log (1+x)}{x \sqrt {1+\sqrt {1+x}}} \, dx=\int \frac {\log (1+x)}{x \sqrt {1+\sqrt {1+x}}} \, dx \]

[In]

Int[Log[1 + x]/(x*Sqrt[1 + Sqrt[1 + x]]),x]

[Out]

Defer[Int][Log[1 + x]/(x*Sqrt[1 + Sqrt[1 + x]]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\log (1+x)}{x \sqrt {1+\sqrt {1+x}}} \, dx \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 0.20 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.07 \[ \int \frac {\log (1+x)}{x \sqrt {1+\sqrt {1+x}}} \, dx=-8 \text {arctanh}\left (\sqrt {1+\sqrt {1+x}}\right )-\frac {2 \log (1+x)}{\sqrt {1+\sqrt {1+x}}}+\frac {\log (1+x) \left (\log \left (\sqrt {2}-\sqrt {1+\sqrt {1+x}}\right )-\log \left (\sqrt {2}+\sqrt {1+\sqrt {1+x}}\right )\right )}{\sqrt {2}}+\sqrt {2} \left (-\log \left (-1+\sqrt {2}\right ) \log \left (1-\sqrt {1+\sqrt {1+x}}\right )+\log \left (1+\sqrt {2}\right ) \log \left (1-\sqrt {1+\sqrt {1+x}}\right )+\log \left (-1+\sqrt {2}\right ) \log \left (1+\sqrt {1+\sqrt {1+x}}\right )-\log \left (1+\sqrt {2}\right ) \log \left (1+\sqrt {1+\sqrt {1+x}}\right )-\operatorname {PolyLog}\left (2,-\left (\left (-1+\sqrt {2}\right ) \left (-1+\sqrt {1+\sqrt {1+x}}\right )\right )\right )+\operatorname {PolyLog}\left (2,\left (1+\sqrt {2}\right ) \left (-1+\sqrt {1+\sqrt {1+x}}\right )\right )+\operatorname {PolyLog}\left (2,\left (-1+\sqrt {2}\right ) \left (1+\sqrt {1+\sqrt {1+x}}\right )\right )-\operatorname {PolyLog}\left (2,-\left (\left (1+\sqrt {2}\right ) \left (1+\sqrt {1+\sqrt {1+x}}\right )\right )\right )\right ) \]

[In]

Integrate[Log[1 + x]/(x*Sqrt[1 + Sqrt[1 + x]]),x]

[Out]

-8*ArcTanh[Sqrt[1 + Sqrt[1 + x]]] - (2*Log[1 + x])/Sqrt[1 + Sqrt[1 + x]] + (Log[1 + x]*(Log[Sqrt[2] - Sqrt[1 +
Sqrt[1 + x]]] - Log[Sqrt[2] + Sqrt[1 + Sqrt[1 + x]]]))/Sqrt[2] + Sqrt[2]*(-(Log[-1 + Sqrt[2]]*Log[1 - Sqrt[1
+ Sqrt[1 + x]]]) + Log[1 + Sqrt[2]]*Log[1 - Sqrt[1 + Sqrt[1 + x]]] + Log[-1 + Sqrt[2]]*Log[1 + Sqrt[1 + Sqrt[1
+ x]]] - Log[1 + Sqrt[2]]*Log[1 + Sqrt[1 + Sqrt[1 + x]]] - PolyLog[2, -((-1 + Sqrt[2])*(-1 + Sqrt[1 + Sqrt[1
+ x]]))] + PolyLog[2, (1 + Sqrt[2])*(-1 + Sqrt[1 + Sqrt[1 + x]])] + PolyLog[2, (-1 + Sqrt[2])*(1 + Sqrt[1 + Sq
rt[1 + x]])] - PolyLog[2, -((1 + Sqrt[2])*(1 + Sqrt[1 + Sqrt[1 + x]]))])

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.29 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.59

method	result	size
derivativedivides	\(8 \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )}{\sum }\frac {\left (\frac {\ln \left (\sqrt {1+\sqrt {1+x}}-\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (1+x \right )}{2}-\operatorname {dilog}\left (\frac {\sqrt {1+\sqrt {1+x}}-1}{-1+\underline {\hspace {1.25 ex}}\alpha }\right )-\ln \left (\sqrt {1+\sqrt {1+x}}-\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {\sqrt {1+\sqrt {1+x}}-1}{-1+\underline {\hspace {1.25 ex}}\alpha }\right )-\operatorname {dilog}\left (\frac {1+\sqrt {1+\sqrt {1+x}}}{1+\underline {\hspace {1.25 ex}}\alpha }\right )-\ln \left (\sqrt {1+\sqrt {1+x}}-\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {1+\sqrt {1+\sqrt {1+x}}}{1+\underline {\hspace {1.25 ex}}\alpha }\right )\right ) \underline {\hspace {1.25 ex}}\alpha }{8}\right )-\frac {2 \ln \left (1+x \right )}{\sqrt {1+\sqrt {1+x}}}-8 \,\operatorname {arctanh}\left (\sqrt {1+\sqrt {1+x}}\right )\)	\(172\)
default	\(8 \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )}{\sum }\frac {\left (\frac {\ln \left (\sqrt {1+\sqrt {1+x}}-\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (1+x \right )}{2}-\operatorname {dilog}\left (\frac {\sqrt {1+\sqrt {1+x}}-1}{-1+\underline {\hspace {1.25 ex}}\alpha }\right )-\ln \left (\sqrt {1+\sqrt {1+x}}-\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {\sqrt {1+\sqrt {1+x}}-1}{-1+\underline {\hspace {1.25 ex}}\alpha }\right )-\operatorname {dilog}\left (\frac {1+\sqrt {1+\sqrt {1+x}}}{1+\underline {\hspace {1.25 ex}}\alpha }\right )-\ln \left (\sqrt {1+\sqrt {1+x}}-\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {1+\sqrt {1+\sqrt {1+x}}}{1+\underline {\hspace {1.25 ex}}\alpha }\right )\right ) \underline {\hspace {1.25 ex}}\alpha }{8}\right )-\frac {2 \ln \left (1+x \right )}{\sqrt {1+\sqrt {1+x}}}-8 \,\operatorname {arctanh}\left (\sqrt {1+\sqrt {1+x}}\right )\)	\(172\)

[In]

int(ln(1+x)/x/(1+(1+x)^(1/2))^(1/2),x,method=_RETURNVERBOSE)

[Out]

8*Sum(1/8*(1/2*ln((1+(1+x)^(1/2))^(1/2)-_alpha)*ln(1+x)-dilog(((1+(1+x)^(1/2))^(1/2)-1)/(-1+_alpha))-ln((1+(1+
x)^(1/2))^(1/2)-_alpha)*ln(((1+(1+x)^(1/2))^(1/2)-1)/(-1+_alpha))-dilog((1+(1+(1+x)^(1/2))^(1/2))/(1+_alpha))-
ln((1+(1+x)^(1/2))^(1/2)-_alpha)*ln((1+(1+(1+x)^(1/2))^(1/2))/(1+_alpha)))*_alpha,_alpha=RootOf(_Z^2-2))-2*ln(
1+x)/(1+(1+x)^(1/2))^(1/2)-8*arctanh((1+(1+x)^(1/2))^(1/2))

Fricas [F(-2)]

Exception generated. \[ \int \frac {\log (1+x)}{x \sqrt {1+\sqrt {1+x}}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(log(1+x)/x/(1+(1+x)^(1/2))^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F(-1)]

Timed out. \[ \int \frac {\log (1+x)}{x \sqrt {1+\sqrt {1+x}}} \, dx=\text {Timed out} \]

[In]

integrate(ln(1+x)/x/(1+(1+x)**(1/2))**(1/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 366, normalized size of antiderivative = 1.26 \[ \int \frac {\log (1+x)}{x \sqrt {1+\sqrt {1+x}}} \, dx=\frac {1}{2} \, {\left (\sqrt {2} \log \left (-\frac {\sqrt {2} - \sqrt {\sqrt {x + 1} + 1}}{\sqrt {2} + \sqrt {\sqrt {x + 1} + 1}}\right ) - \frac {4}{\sqrt {\sqrt {x + 1} + 1}}\right )} \log \left (x + 1\right ) + \sqrt {2} {\left (\log \left (\sqrt {2} + \sqrt {\sqrt {x + 1} + 1}\right ) \log \left (-\frac {\sqrt {2} + \sqrt {\sqrt {x + 1} + 1}}{\sqrt {2} + 1} + 1\right ) + {\rm Li}_2\left (\frac {\sqrt {2} + \sqrt {\sqrt {x + 1} + 1}}{\sqrt {2} + 1}\right )\right )} - \sqrt {2} {\left (\log \left (-\sqrt {2} + \sqrt {\sqrt {x + 1} + 1}\right ) \log \left (-\frac {\sqrt {2} - \sqrt {\sqrt {x + 1} + 1}}{\sqrt {2} + 1} + 1\right ) + {\rm Li}_2\left (\frac {\sqrt {2} - \sqrt {\sqrt {x + 1} + 1}}{\sqrt {2} + 1}\right )\right )} + \sqrt {2} {\left (\log \left (\sqrt {2} + \sqrt {\sqrt {x + 1} + 1}\right ) \log \left (-\frac {\sqrt {2} + \sqrt {\sqrt {x + 1} + 1}}{\sqrt {2} - 1} + 1\right ) + {\rm Li}_2\left (\frac {\sqrt {2} + \sqrt {\sqrt {x + 1} + 1}}{\sqrt {2} - 1}\right )\right )} - \sqrt {2} {\left (\log \left (-\sqrt {2} + \sqrt {\sqrt {x + 1} + 1}\right ) \log \left (-\frac {\sqrt {2} - \sqrt {\sqrt {x + 1} + 1}}{\sqrt {2} - 1} + 1\right ) + {\rm Li}_2\left (\frac {\sqrt {2} - \sqrt {\sqrt {x + 1} + 1}}{\sqrt {2} - 1}\right )\right )} - 4 \, \log \left (\sqrt {\sqrt {x + 1} + 1} + 1\right ) + 4 \, \log \left (\sqrt {\sqrt {x + 1} + 1} - 1\right ) \]

[In]

integrate(log(1+x)/x/(1+(1+x)^(1/2))^(1/2),x, algorithm="maxima")

[Out]

1/2*(sqrt(2)*log(-(sqrt(2) - sqrt(sqrt(x + 1) + 1))/(sqrt(2) + sqrt(sqrt(x + 1) + 1))) - 4/sqrt(sqrt(x + 1) +
1))*log(x + 1) + sqrt(2)*(log(sqrt(2) + sqrt(sqrt(x + 1) + 1))*log(-(sqrt(2) + sqrt(sqrt(x + 1) + 1))/(sqrt(2)
+ 1) + 1) + dilog((sqrt(2) + sqrt(sqrt(x + 1) + 1))/(sqrt(2) + 1))) - sqrt(2)*(log(-sqrt(2) + sqrt(sqrt(x + 1
) + 1))*log(-(sqrt(2) - sqrt(sqrt(x + 1) + 1))/(sqrt(2) + 1) + 1) + dilog((sqrt(2) - sqrt(sqrt(x + 1) + 1))/(s
qrt(2) + 1))) + sqrt(2)*(log(sqrt(2) + sqrt(sqrt(x + 1) + 1))*log(-(sqrt(2) + sqrt(sqrt(x + 1) + 1))/(sqrt(2)
- 1) + 1) + dilog((sqrt(2) + sqrt(sqrt(x + 1) + 1))/(sqrt(2) - 1))) - sqrt(2)*(log(-sqrt(2) + sqrt(sqrt(x + 1)
+ 1))*log(-(sqrt(2) - sqrt(sqrt(x + 1) + 1))/(sqrt(2) - 1) + 1) + dilog((sqrt(2) - sqrt(sqrt(x + 1) + 1))/(sq
rt(2) - 1))) - 4*log(sqrt(sqrt(x + 1) + 1) + 1) + 4*log(sqrt(sqrt(x + 1) + 1) - 1)

Giac [F]

\[ \int \frac {\log (1+x)}{x \sqrt {1+\sqrt {1+x}}} \, dx=\int { \frac {\log \left (x + 1\right )}{x \sqrt {\sqrt {x + 1} + 1}} \,d x } \]

[In]

integrate(log(1+x)/x/(1+(1+x)^(1/2))^(1/2),x, algorithm="giac")

[Out]

integrate(log(x + 1)/(x*sqrt(sqrt(x + 1) + 1)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\log (1+x)}{x \sqrt {1+\sqrt {1+x}}} \, dx=\int \frac {\ln \left (x+1\right )}{x\,\sqrt {\sqrt {x+1}+1}} \,d x \]

[In]

int(log(x + 1)/(x*((x + 1)^(1/2) + 1)^(1/2)),x)

[Out]

int(log(x + 1)/(x*((x + 1)^(1/2) + 1)^(1/2)), x)